Anyway my problem is I only know how to define n! as a recursive definition, so I'm not really sure how to demonstrate this truth in any kind of mathematical way. If you put both as a reiman product or whatever it's called (ie n-i for i=0 to n and 2 for i=0 to n) you have a function compared to a number.

like

2*2*2*2*....*2 (n terms)
1*2*...*(n-1)*(n) (n terms)

I don't really wanna use induction since I dont think it's necessary or expected, and I don't really know how to do the inductive step since one of the functions has no variables.

Anyway any ideas? Or anyone know a non recursive definition for factorial so I can compare it to 2^n? heh

But don't both have variables? Consider a function h(x) = g(x)/f(x) = 2^x/x!

isn't h(x+1) just h(x) * 2/(x+1) ? That would be your induction.. h(x+1)/h(x) obviously converges to 0 as x approaches infinity so you know that the denominator is growing faster than the numerator of h(x) as x gets very large.

Edit: Sorry I don't know what level of class this is for so I don't know whether to make this proof look nicer or not. Tell me if this is not what you are looking for.

Oh so you're trying to do something with Riemann product notation? (pi?) Edit I see why you're saying there's no variable attached. Heh the only thing I can think of is that (n+1)! is (n+1)*pi(i, for i = 1 to n) and 2^(n+1) being 2 * pi(2, for i = 1 to n) but I don't get why this is such an impediment.

Edit again: ok so I defined h(x) to be 2^x / x! right?

h(x+1) can be expressed as h(x)*2/(x+1) because

h(x)*2/(x+1) = 2^x/x! * 2/(x+1)

= 2 * 2^x / (x! * (x+1))

= 2^(x+1)/(x+1)!

that h(x+1) = h(x) * 2/(x+1) would be sufficient to prove x! increases faster than 2^x but you keep indicating that induction is too complicated for the question and that there is a simpler way. Sorry I can't see one :\